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3 years ago

What is a force?

Physics

1 answer:

Fudgin [204]3 years ago

5 0

Answer:

2. A push or pull

Explanation:

I'm 100% sure

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A fishbowl has a circular opening with a diameter of 15 cm. The fishbowl sits upright on a table in a magnetic field of 0.00110

givi [52]

Answer:

−1.9 × 10−5 Wb

Explanation:

Magnetic flux is the product of average magnetic field times the perpendicular area that is crosses times the angle between planar area and magnetic flux

Finding the area

Area = πr² = 3.14*0.075*0.075

Area=0.01767 m²

Magnetic flux = 0.01767*0.00110*cos 0

Magnetic flux =0.000019 = −1.9 × 10−5 Wb

7 0

3 years ago

Think about the electricity sent to your home from a power plant. How does the voltage of the electricity that leaves the plant

Lana71 [14]

The voltage at the plant is HUGEEEE compared to the one at your house. your house functions at 230V , the plant works at thousands times more voltage

3 0

3 years ago

Read 2 more answers

Let r denote the distance between the center of the earth and the center of the moon. What is the magnitude of the acceleration

stepan [7]

Answer:

The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is $\mathbf{3.3187\times10^{-5}}\frac{m}{s^{2}}$

Explanation:

By Newton's gravitational law, the magnitude of the gravitational force between two objects is:

$F=G\frac{Mm}{r^{2}}$ (1)

With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:

$M=5.972\times10^{24}\,kg$

$m=7.34767309\times10^{22}\,kg$

$r=384400\,km$

$G=6.674\times10^{-11}\,\frac{N\,m^{3}}{kg^{2}}$

Using those values on (1)

$F=(6.674\times10^{-11})*\frac{(5.972\times10^{24})(7.34767309\times10^{22})}{(384400\times10^{3})^{2}}$

$F\approx1.98193\times10^{20}N$

Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:

$F=M*ae\Longrightarrow ae=\frac{F}{M}=\frac{1.98193\times10^{20}}{5.972\times10^{24}}\approx\mathbf{3.3187\times10^{-5}}$

6 0

4 years ago

Calculate the kinetic energy of a 0.032 kg ball as it leaves a hand to be thrown upwards at 6.2 m/s

AnnZ [28]

Answer:

The ball will have a kinetic energy of 0.615 Joules.

Explanation:

Use the kinetic energy formula

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}0.032kg\cdot 6.2^2 \frac{m^2}{s^2}= 0.615J$

The kinetic energy at the moment of leaving the hand will be 0.615 Joules. (From there on, as it ball is traveling upwards, this energy will be gradually traded off with potential energy until the ball's velocity becomes zero at the apex of the flight)

3 0

4 years ago

A 30000 grams boy is riding a merry-go-round with a radius of 600 cm. What is the centripetal force and acceleration on the boy

irga5000 [103]

Answer:

centripetal force is calculated by mass(kg) × tangetial velocity(m/s) ÷ radius (m)

Explanation:

so 30000g= 30kg

50km/h = 13.88m/s

600cm= 6m

30×13.88÷6= 69.4N

N= Newton's

hope this helps.

btw I'm 16 and love physics so I tried my best in this hope it went well!!

7 0

3 years ago